(2) 1/(x+3)>0 to obtain x & gt-3.

The definition domain of the inverse function is the original function value domain, so find the original function value domain first.

(3)x & gt； 1/3

3x- 1 >0

y∈R

y/2=log2(3x- 1)

2^(y/2)=3x- 1

x=[2^(y/2)+ 1]/3

The inverse function is y = [2 (x/2)+ 1]/3, x ∈ r.

(4) let u = 2 x, u >；; 0

y = u/(u+ 1)=[(u+ 1)- 1]/(u+ 1)= 1- 1/(u+ 1)

U>0 and then u+1>; 1，0 & lt； 1/(u+ 1)& lt； 1，- 1 & lt； - 1/(u+ 1)& lt； 0，0 & lt； y & lt 1

y=u/(u+ 1)

yu+y=u

u=y/( 1-y)

2^x=y/( 1-y)

X = log (2) [y/(1-y)] (base 2)

y=log(2)[x/( 1-x)] x∈(0， 1)

(5)y=u^2,u=2x- 1

(6)y=e^u,u=3x+8

(7)y= cube root number u, u = v 2, v = 1-lnx.

(8)y=u^3,u=arcsinv,v=ax+b

(9) Simultaneous three inequalities: x 2-1≠ 0,-1≤x≤ 1, and x≥0.

0 ≤ x